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Statistic Homework Sheet…MUST HAVE EXCEL SHEET. NO HANDWRITTEN WORK ACCEPTED

Module 05 Homework assignment

Name:______________________________________
Module 5 Homework Assignment
The paired data below consists of test scores and hours of preparation for 5 randomly selected
students. Use this data set to answer the questions below:
x Hours of preparation 5 2 9 6 10
y Test score
64 48 72 73 80
1. Use the given data to find the correlation coefficient r, regression equation and scatter plot in MS
Excel.
Solution:
2. Based on the linear correlation coefficient r, is this a good model? Explain.
Solution:
3. What is the best predicted test score for a student who spent 7 hours preparing for the test?
Solution:
4. Find the standard error se . Use formula or MS Excel.
Solution:
5. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the
test given that E=34.677. Interpret the result.
Solution:
6. Find the explained variation.
Solution:
7. Find the unexplained variation.
Solution:
8. Find the total variation.
Solution:
9. Find the value of r2 and explain its meaning.
Solution:
10. If the data point ( 3, 100) is added to the data set, how would this effect the results of the regression
analysis? Is this data point an outlier, influential point or both? Explain.
Solution:
Name:______________________________________
Module 5 Homework Assignment
The paired data below consists of test scores and hours of preparation for 5 randomly selected
students. Use this data set to answer the questions below:
x Hours of preparation 6 3 8 5 9
y Test score
69 50 68 65 80
1. Use the given data to find the correlation coefficient r, regression equation and scatter plot in MS
Excel.
Solution:
Instructor
Scatter Plot:
Hours of Study v Test Score
90
80
70
60
50
40
30
20
10
0
0
2
4
6
8
10
Correlation coefficient uses the Excel formula =CORREL(y-values, x-values)
Ã°Ââ€˜Å¸ = 0.9186
Regression equation:
Ã°Ââ€˜Â¦ÃŒâ€š = Ã°Ââ€˜Â0 + Ã°Ââ€˜Â1 Ã°Ââ€˜Â¥
The y-intercept of the regression equation (Ã°Ââ€˜Â0 ) is found using the Excel formula
=INTERCEPT(y-values, x-values).
The slope of the regression equation (Ã°Ââ€˜Â1 ) is found using the Excel formula
=SLOPE(y-values, x-values).
Ã°Ââ€˜Â¦ÃŒâ€š = 40.6754 + 4.1491Ã°Ââ€˜Â¥
2. Based on the linear correlation coefficient r, is this a good model? Explain.
Solution:
This question depends on what value of alpha we
choose. LetÃ¢â‚¬â„¢s see what happens with both Ã°Ââ€ºÂ¼ =
0.05 and Ã°Ââ€ºÂ¼ = 0.01:
Ã°Ââ€ºÂ¼ = 0.05:
Using the table of critical values of Ã°Ââ€˜Å¸ from page
692 of the textbook we get a critical value of
0.878 for a sample size of 5. Since the correlation
coefficient is greater than the critical value we
conclude that there is evidence of linear
correlation between hours of study and test
scores and therefore the regression equation is a
good model and can be used for predicting test
scores based of hours of study.
Ã°Ââ€ºÂ¼ = 0.01:
Using the table of critical values of Ã°Ââ€˜Å¸ from page
692 of the textbook we get a critical value of
0.959 for a sample size of 5. Since the correlation
coefficient is less than the critical value we
conclude that there is not evidence of linear
correlation between hours of study and test
scores, and therefore the regression equation is
not a good model and cannot be used for
predicting test scores based of hours of study.
3. What is the best predicted test score for a student who spent 4 hours preparing for the test?
Solution:
Again this question depends on what value of
alpha we choose. LetÃ¢â‚¬â„¢s see what happens with
both Ã°Ââ€ºÂ¼ = 0.05 and Ã°Ââ€ºÂ¼ = 0.01:
Ã°Ââ€ºÂ¼ = 0.05:
At this level of significance we are OK with using
the regression equation to predict test scores, so
we plug in Ã°Ââ€˜Â¥ = 4 into the regression equation to
get the predicted value Ã°Ââ€˜Â¦ÃŒâ€š:
Ã°Ââ€˜Â¦ÃŒâ€š = 40.6754 + 4.1491(4)
Ã°Ââ€˜Â¦ÃŒâ€š = 57.2719
Ã°Ââ€ºÂ¼ = 0.01:
At this significance level we are saying that there
is no correlation between hours of study and test
scores, so the regression equation is not used. In
this case our best predicted value is simply the
average of the test scores, Ã°Ââ€˜Â¦ÃŒâ€¦:
Ã°Ââ€˜Â¦ÃŒâ€¦ = 66.4
4. Find the standard error se . Use formula or MS Excel.
Solution:
Use the Regression command from the Data
Analysis ToolPak add-in for Excel to find the
standard error (as well as explained variation,
unexplained variation, and total variation from
#Ã¢â‚¬â„¢s 6, 7, and 8).
Ã°Ââ€˜Â Ã°Ââ€˜â€™
5. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the
test given that E=34.677. Interpret the result.
Solution: